If `3(a+2c)=4(b+3d),` then the equation `a x^3+b x^2+c x+d=0` will have no real solution at least one real root in `(-1,0)` at least one real root in `(0,1)` none of these

Options

A. no real solution
B. Atleast one real root in `(-1,0)`
C. Atleast on real root in `(0,1)`
D. none of these

Answer

B

Explanation

Let `f(x)=(ax^(4))/4+(bx^(3))/3+(cx^(2))/2+dx,`
Which is continuous and differentiable `f(0)=0`
`f(-1)=1/4-b/3+c/2-d=1/4(a+2c)=1/3(b+3d)=0`
So, according to Rolle's theorem, there exists atleast one root of `f^(')(x)=0` in `(-1,0)`